To determine the bond order, we first need to write the molecular orbital (MO) electronic configuration for each species and then apply the bond order formula:

$\text{Bond Order}=\frac{1}{2}(\text{Number of electrons in bonding MOs}-\text{Number of electrons in antibonding MOs})$

Step 1: Determine total electrons for each species.

• ${\mathrm{O}}_2^{2-}$: $2\times 8+2=18$ electrons
• ${\mathrm{O}}_2$: $2\times 8=16$ electrons
• ${\mathrm{O}}_2{}^{+}$: $2\times 8-1=15$ electrons
• ${\mathrm{O}}_2^{-}$: $2\times 8+1=17$ electrons

Step 2: Write MO configuration and calculate bond order.

The MO configuration for species with more than 14 electrons follows the order:

$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2(\pi 2p_x^2=\pi 2p_y^2)({\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1){\sigma }^{\ast }2p_z^0$

• A) ${\mathrm{O}}_2^{2-}$ (18 electrons):
  $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^2{\pi }^{\ast }2p_y^2$
  Bonding electrons = 10, Antibonding electrons = 8
  Bond Order = $\frac{1}{2}(10-8)=1$
• B) ${\mathrm{O}}_2$ (16 electrons):
  $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^1{\pi }^{\ast }2p_y^1$
  Bonding electrons = 10, Antibonding electrons = 6
  Bond Order = $\frac{1}{2}(10-6)=2$
• C) ${\mathrm{O}}_2{}^{+}$ (15 electrons):
  $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^1$
  Bonding electrons = 10, Antibonding electrons = 5
  Bond Order = $\frac{1}{2}(10-5)=2.5$
• D) ${\mathrm{O}}_2^{-}$ (17 electrons):
  $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^2{\pi }^{\ast }2p_y^1$
  Bonding electrons = 10, Antibonding electrons = 7
  Bond Order = $\frac{1}{2}(10-7)=1.5$

Therefore, ${\mathrm{O}}_2^{-}$ has a bond order of 1.5.

Correct Answer: (D)